We are given that in an office, 6outof10 employees were interested in doing work from home.

If three employees were selected at random from the above 10 employees, then we need to find the probability that at most two of the selected were interested in doing work from home.

Because replacement is not mentioned then by default, we will consider without replacement and we will use combinations.

Now, the total number of ways of selecting three employees out of ten employees in the office is 10𝐶3.
Since 6 out of 10employees were interested in doing work from home, hence, for the probability of selecting at most two of them from these, we will subtract from 1 the probability of selecting all the three employees interested in doing the work from home.
Using the formula; 𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 = 𝐹𝑎𝑣𝑜𝑟𝑎𝑏𝑙𝑒 𝐶𝑎𝑠𝑒𝑠 𝑇𝑜𝑡𝑎𝑙 𝐶𝑎𝑠𝑒𝑠
Probability of selecting 3 employees out of 6 employees interested in
6𝐶3 doing the work from home is 10𝐶3 Required probability will be 5

Probability
6𝐶 ( 6! )
=1− 3 =1− 3!×3! 10𝐶3 ( 10! )
3! × 7!
=1− 6×5×4 =1−1 10×9×8 6
=5/6
The correct answer is Choice D.

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